Chapter 9.6, Problem 1BGP

To determine

Find the solution.

Answer to Problem 1BGP

  $x=-0.33$ or $x=2$

Explanation of Solution

Given Information: $-3x^2+5x=-2$

Formula Used: $a^2-2ab+b^2={(a-b)}^2$ and $a^2-b^2=(a+b)(a-b)$

Calculation:

  $-3x^2+5x=-2$

So, simplify

  $3x^2-5x-2=0$

Now, to make coefficient of $x^2$ = 1, divide the equation by 3.

  $x^2-\frac{5}{3}x-\frac{2}{3}=0$

Now, adding & subtracting the square of half the coefficient of x.

  $x^2-\frac{5}{3}x-\frac{2}{3}+{\left(\frac{5}{6}\right)}^2-{\left(\frac{5}{6}\right)}^2=0$

Now use formula, $a^2-2ab+b^2={(a-b)}^2$

  ${\left(x-\frac{5}{6}\right)}^2-\frac{2}{3}-{\left(\frac{5}{6}\right)}^2=0$

So,

  ${\left(x-\frac{5}{6}\right)}^2-\frac{2}{3}-\frac{25}{36}=0$

Now,

  ${\left(x-\frac{5}{6}\right)}^2-\frac{24-25}{36}=0$

Now,

  ${\left(x-\frac{5}{6}\right)}^2-\frac{49}{36}=0$

Now, square root of 49 is 7 and square root of 36 is 6.

  ${\left(x-\frac{5}{6}\right)}^2-{\left(\frac{7}{6}\right)}^2=0$

Now use formula, $a^2-b^2=(a+b)(a-b)$

  $\left(x-\frac{5}{6}+\frac{7}{6}\right)\left(x-\frac{5}{6}-\frac{7}{6}\right)=0$

Now,

  $x=\frac{5}{6}-\frac{7}{6}$ or $x=\frac{5}{6}+\frac{7}{6}$

Now,

  $x=\frac{5-7}{6}$ or $x=\frac{5+7}{6}$

Now,

  $x=\frac{-2}{6}$ or $x=\frac{12}{6}$

Hence,

  $x=-0.33$ or $x=2$